Question

表：购物

shop_id shop_building shop_person  shop_time
1   1   Brian   40
2   2   Brian   31
3   1   Tom    20
4   3   Brian   30

表：建筑物

building_id building_city
1     London
2     Newcastle 
3     London
4     London

表：香蕉

banana_id  banana_building banana_amount  banana_person
1      2     1      Brian
2      3     1       Brian
2      1     1      Tom

我现在希望它向我展示每个人在伦敦购买的香蕉数量。

我使用了这段代码：

SELECT tt.*, tu.*, tz.*,
           SUM(shop_time)           AS shoptime, 
           Ifnull(banana_amount, 0) AS bananas 
    INNER JOIN buildings tu ON tt.shop_building=tu.building_id
    FROM   shopping tt 
           LEFT OUTER JOIN (SELECT banana_person, banana_building,
                                   SUM(banana_amount) AS banana_amount 
                            FROM   bananas 
                            GROUP  BY banana_person) tz 
             ON tt.shop_person = tz.banana_person AND tt.shop_building = tz.banana_building
 WHERE tu.building_city = 'London'
    GROUP  BY shop_person;

但它不起作用。这就好像我告诉它为时已晚，它应该只在伦敦看，因为它忽略了这一点。

Answer 1

尝试这种方式：

SELECT 
    s.shop_person, sum(b.banana_amount) as Amt, , sum(shop_time) as TimeAmt
FROM bananas b
    INNER JOIN buildings bu ON b.banana_building = bu.building_id
    INNER JOIN shopping s ON bu.building_id = s.shop_building
WHERE
    bu.building_city = N'London'
GROUP BY s.shop_person

这个查询不同，但它可以做你想要的 - “每个人在伦敦购买的香蕉数量”

Answer 2

在不知道您使用的数据库的情况下，这是mssql的工作版本。我实际上重建了表格以确保它是正确的。

对于其他数据库系统，您可能必须在SELECT语句中使用除ISNULL之外的其他函数。

SELECT tt.shop_person,
    tt.shop_building,
    SUM(tt.shop_time) AS shoptime,
    ISNULL(SUM(tz.banana_amount), 0) AS bananas
FROM dbo.shopping tt
INNER JOIN dbo.buildings tu ON tt.shop_building = tu.building_id
LEFT OUTER JOIN
    (SELECT banana_person, banana_building, SUM(banana_amount) AS banana_amount
        FROM bananas
        GROUP BY banana_person, banana_building) tz
    ON tt.shop_person = tz.banana_person AND tt.shop_building = tz.banana_building
WHERE (tu.building_city = 'London')
GROUP BY tt.shop_person, tt.shop_building

我必须在tz.banana_amount附近添加一个聚合函数 - 其中一个（SUM，MIN，MAX）无关紧要。

结果：

shop_person shop_building shoptime bananas
Brian       1             40       0
Tom         1             20       1
Brian       3             30       1

我在bananas等中玩了不同数量的游戏，但它运行正常。

交叉连接忽略where子句

2 个答案: